# BREAKING DOWN THE YEAR

This topic is a little math intensive and is not necessary for you to understand the rest of this chapter. The only sophisticated math, however, is handled using a spread­sheet function. If you’re willing to “go with” the use of the spreadsheet function and a brief explanation of what’ s happening, this section will give you a little more insight into the workings of Life Tables.

For whatever reason, an insurance company decided it would like to be able to sell term policies for half years rather than years. In order to price these policies, it

Table 10.6 Excerpt from the Life Table for Men with Some Curve Fitting Data

 Age q i Number of dead By fit % Error 45 0.003735 94,154 5,846 5,854 -0.14 46 0.004071 93,803 6,197 6,196 0.02 47 0.004428 93,421 6,579 6,574 0.07 48 0.004806 93,007 6,993 6,989 0.06 49 0.005206 92,560 7,440 7,439 0.01 50 0.005648 92,078 7,922 7,926 -0.06 51 0.006121 91,558 8,442 8,449 -0.09 52 0.006594 90,998 9,002 9,008 -0.07 53 0.007045 90,398 9,602 9,604 -0.01 54 0.007488 89,761 10,239 10,235 0.04

needed Life Tables with half-year rather than full-year steps. For a crude approxima­tion, it could have just split the q yearly values in half, for example, for a man, q50 = 0.005648, so it could have used 0.005648/2 = 0.002824 for the first half of his fiftieth year and the same number for the second half. This is not a terrible approxi­mation, but since the q values are climbing year by year in the 50-year age range, the result would have been to overcharge for a first – half fiftieth year policy and to undercharge for a second-half fiftieth year policy.

My first step in creating a more accurate set of q values is to create a notation. The probability of a man dying sometime during his fiftieth year is q50. I’ll call the new set of numbers r to distinguish these numbers from the q numbers. Remember that there will be an r number for the first half of each year’s age and a different r number for the second half.

Table 10.6 starts with an excerpt from the men’s Life Table. I’m only consider­ing ages 45-54 and I’m only copying the age, q, and – columns. Remember that column l is the number of men alive out of the original 100,000. In the next column, I showed the number of men dead out of the original 100,000. This column was generated by simply subtracting the numbers in column l from 100,000. Figure 10.6 shows a plot of the number of men dead column versus the man’s age.

Spreadsheet programs offer the ability to “fit” a formula to a set of data. This means that the spreadsheet program takes a formula that you choose from a list of available formulas and adjusts the parameters in this formula to make a plot of it look as though this formula had actually created the original data. Some art is involved in the choice of formula. For the data shown in Figure 10.6, I chose a “second – order polynomial” as my formula. This is the part you don ’ t have to worry about if you’re uncomfortable with the math, but keep reading for the results.

The spreadsheet program calculated the formula

Number of dead = 18.09age2 – 1,304.1age + 27,906

 5,000 45 47 49 51 53 55 Age (years) Figure 10.6 Curve fit used to generate a half-year Life Table section.

as its best fit to the data. In Table 10.6, I showed the formula predictions for the number of dead (labeled “by fit”) and the percent error in the formula prediction versus the original data. In Figure 10.6, I have superimposed a graph of the formula on top of the graph of the original data. As you can see from the table, the percent error is very small. And as you can see from the figure, it’s just about impossible to tell the difference between the two.

Table 10.7 shows how I built my new life table using my formula. The first column lists ages in half-year steps. The second column lists the number of dead generated by using my formula. The third column lists the number of alive (,), generated by subtracting each number of dead value from the original 100,000. The fourth column is the Life Table d column, the number of men dying each year. This is generated by subtracting the number of alive in one row by the number of alive in the previous row, for example, d45.5 = l23 – 124.

Looking at the original Life Tables, d values are obtained by multiplying the number of people alive by the probability of their dying, dt = ql where i refers to any row (age) in the table. To get my new probability values, r, therefore, I just divide di by li. For example, r45.5 = d45.5/l45.5 = 176/93,980 = 0.001769.[28]

The insurance company would no doubt do much more exotic curve fitting than I did and extend its results over the entire 0- to 100-year age span, but what I did above shows the calculations involved. I now have a Life Table that can accurately be used to prepare term policies in half-year increments.

Table 10.7 Half Year Life Table for Men

 Age Number of dead l d r 45.0 5,854 94,146 167 0.001769 45.5 6,020 93,980 176 0.001868 46.0 6,196 93,804 185 0.001968 46.5 6,380 93,620 194 0.002069 47.0 6,574 93,426 203 0.002170 47.5 6,777 93,223 212 0.002271 48.0 6,989 93,011 221 0.002374 48.5 7,209 92,791 230 0.002477 49.0 7,439 92,561 239 0.002581 49.5 7,678 92,322 248 0.002685 50.0 7,926 92,074 257 0.002791 50.5 8,183 91,817 266 0.002897 51.0 8,449 91,551 275 0.003004 51.5 8,724 91,276 284 0.003113 52.0 9,008 90,992 293 0.003222 52.5 9,301 90,699 302 0.003332 53.0 9,604 90,396 311 0.003443 53.5 9,915 90,085 320 0.003555 54.0 10,235 89,765 329 0.003669 54.5 10,564 89,436 338 0.003783

PROBLEMS

1. A-50-year old man decides that he wants a \$1,000,000 life insurance for 2 years.

(a) What will two separate 1-year term policies cost, both bought at the beginning of the first year? Use 4% annual percentage rate (APR) for calculating present values and ignore insurance company markups.

(b) Repeat the above but treat this as a 2-year policy bought on the man’s fiftieth birthday.

2. Suppose someone was to come down with an incurable terminal disease and was told by his or her doctor that “You have a two-third chance of lasting a year; you’ll never last more than 2 years.” Construct a Life Table for this person and calculate the cost of a \$100,000 life insurance policy for this person.

3. Using the half year Life Table shown in Table 10.6, assuming an APR of 4% for the present value, what is the cost (idealized) of an 18-month, \$50,000 term policy for a man starting halfway through his forty – eighth year?

4. Life Tables can be generated for longer time steps than 1 year as well as for shorter time steps than 1 year. Tables generated in 5-year steps are sometimes called abridged Life Tables. Starting with the women’s Life Table, generate an abridged Women’s Life Table.

5. You are taking a 5-year business loan for \$250,000 that you will amortize yearly with 60 equal monthly payments at an APR of 10%. Your lender wants a fully paid-up life insur­ance policy to protect his or her interest. You will get a decreasing term policy that pays exactly the balance at the beginning of each year. You are a healthy, 35-year-old male.

(a) Create an amortization table for the loan and show the balances at the beginning of each year.

(b) Using the balances at the beginning of each year from problem 5a, create the 5-year decreasing term policy and get the price for this policy as an up-front payment. Use 5% as the APR for calculating present values. Assume that your insurance company prices this policy at 35% above the ideal calculation number.

(c) If you borrow the money for the policy along with your loan, what’s your effective APR (effective annual percentage rate [EAPR])?